Jim Lambers
MAT 280
Spring Semester 2009-10
Lecture 14 Notes
These notes correspond to Sections 12.6 and 12.7 in Stewart and Sections 1.4 and 6.2 in Marsden
and Tromba.
Triple Integrals in Cylindrical Coordinates
We have seen that in some cases, it is convenient to evaluate double integrals by converting Cartesian
coordinates (𝑥, 𝑦) to polar coordinates (𝑟, 𝜃). The same is true of triple integrals. When this is the
case, Cartesian coordinates (𝑥, 𝑦, 𝑧) are converted to cylindrical coordinates (𝑟, 𝜃, 𝑧).
The relationships between (𝑥, 𝑦) and (𝑟, 𝜃) are exactly the same as in polar coordinates, and
the 𝑧 coordinate is unchanged.
Example The point (𝑥, 𝑦, 𝑧) = (−3, 3, 4) can be converted to cylindrical coordinates (𝑟, 𝜃, 𝑧) using
the relationships from polar coordinates,
√
𝑦
𝑟 = 𝑥2 + 𝑦 2 , tan 𝜃 = .
𝑥
These relationships yield
𝑟=
√
√
√
32 + (−3)2 = 18 = 3 2,
tan 𝜃 = −1.
Since 𝑥 = −3 < 0, we have 𝜃 √
= tan−1 (−1)+𝜋 = 3𝜋/4. We conclude that the cylindrical coordinates
of the point (−3, 3, 4) are (3 2, 3𝜋/4, 4). □
Furthermore, just as conversion to polar coordinates in double integrals introduces a factor of
𝑟 in the integrand, conversion to cylindrical coordinates in triple integrals also introduces a factor
of 𝑟.
Example We evaluate the triple integral
∫ ∫ ∫
𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉,
𝐸
where 𝐸 is the solid bounded below by the paraboloid 𝑧 = 𝑥2 + 𝑦 2 , above by the plane 𝑧 = 4, and
the planes 𝑦 = 0 and 𝑦 = 2. This integral can be evaluated as an iterated integral
∫
2
√
∫
4−𝑥2
∫
4
𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥,
−2
0
𝑥2 +𝑦 2
1
but if we instead describe the region using cylindrical coordinates, we find that the solid is bounded
below by the paraboloid 𝑧 = 𝑟2 , above by the plane 𝑧 = 4, and contained within the polar “box”
0 ≤ 𝑟 ≤ 2, 0 ≤ 𝜃 ≤ 𝜋. We can therefore evaluate the iterated integral
∫ 2∫ 𝜋∫ 4
𝑓 (𝑟 cos 𝜃, 𝑟 sin 𝜃, 𝑧) 𝑟 𝑑𝑧 𝑑𝜃 𝑑𝑟,
0
0
𝑟2
that has much simpler limits. □
Triple Integrals in Spherical Coordinates
Another approach to evaluating triple integrals, that is especially useful when integrating over
regions that are at least partially defined using spheres, is to use spherical coordinates. Consider a
point (𝑥, 𝑦, 𝑧) that lies on a sphere of radius 𝜌. Then we know that 𝑥2 + 𝑦 2 + 𝑧 2 = 𝜌2 . Furthermore,
the
√ points (0, 0, 0), (0, 0, 𝑧) and (𝑥, 𝑦, 𝑧) form a right triangle with hypotenuse 𝜌 and legs ∣𝑧∣ and
𝜌2 − 𝑧 2 .
If we denote by 𝜙 the angle adjacent to the leg of length ∣𝑧∣, then 𝜙 can be interpreted as an
angle of inclination of the point (𝑥, 𝑦, 𝑧). The angle 𝜙 = 0 corresonds to the “north pole” of the
sphere, while 𝜙 = 𝜋/2 corresponds to the “equator”, and 𝜙 = 𝜋 corresponds to the “south pole”.
By right triangle trigonometry, we have
𝑧 = 𝜌 cos 𝜙.
It follows that 𝑥2 + 𝑦 2 = 𝜌2 sin2 𝜙. If we define the angle 𝜃 to have the same meaning as in polar
coordinates, then we have
𝑥 = 𝜌 sin 𝜙 cos 𝜃, 𝑦 = 𝜌 sin 𝜙 sin 𝜃.
We define the spherical coordinates of (𝑥, 𝑦, 𝑧) to be (𝜌, 𝜃, 𝜙).
√
Example To convert the point (𝑥, 𝑦, 𝑧) = (1, 3, −4) to spherical coordinates, we first compute
√
√
√
√
√
2
2
2
𝜌 = 𝑥 + 𝑦 + 𝑧 = 12 + ( 3)2 + (−4)2 = 20 = 2 5.
Next, we use the relation tan 𝜃 = 𝑦/𝑥, and the fact that 𝑥 = 1 > 0, to obtain
𝜃 = tan−1
√
𝑦
𝜋
= tan−1 3 = .
𝑥
3
Finally, to obtain 𝜙, we use the relation 𝑧 = 𝜌 cos 𝜙, which yields
(
)
4
−1 𝑧
−1
𝜙 = cos
= cos
− √
≈ 2.6779 radians.
𝜌
2 5
□
2
To evaluate integrals in spherical coordinates, it is important to note that the volume of a
“spherical box” of dimensions Δ𝑟, Δ𝜃 and Δ𝜙, as Δ𝜌, Δ𝜃, Δ𝜙 → 0, converges to the infinitesimal
𝜌2 sin 𝜙 𝑑𝑟 𝑑𝜃 𝑑𝜙,
where (𝜌, 𝜃, 𝜙) denotes the location of the box in the limit. Therefore, the integral of a function
𝑓 (𝑥, 𝑦, 𝑧) over a solid 𝐸, when evaluated in spherical coordinates, becomes
∫ ∫ ∫
∫ ∫ ∫
𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 =
𝑓 (𝜌 sin 𝜙 cos 𝜃, 𝜌 sin 𝜙 sin 𝜃, 𝜌 cos 𝜙) 𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜃 𝑑𝜙.
𝐸
𝐸
Example We wish to compute the volume of the solid 𝐸 in the first octant bounded below by the
plane 𝑧 = 0 and the hemisphere 𝑥2 +𝑦 2 +𝑧 2 = 9, bounded above by the hemisphere 𝑥2 +𝑦 2 +𝑧 2 = 16,
and the planes 𝑦 = 0 and 𝑦 = 𝑥. This would be highly inconvenient to attempt to evaluate in
Cartesian coordinates; determining the limits in 𝑧 alone requires breaking up the integral with
respect to 𝑧. However, in spherical coordinates, the solid 𝐸 is determined by the inequalities
3 ≤ 𝜌 ≤ 4,
𝜋
,
4
0≤𝜃≤
0≤𝜙≤
𝜋
.
2
That is, the solid is actually a “spherical rectangle”. It follows that the volume 𝑉 is given by the
iterated integral
∫
𝑉
𝜋/2 ∫ 𝜋/4 ∫ 4
=
0
=
=
=
𝜋
4
∫
𝜋
4
∫
𝜋
4
∫
𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜃 𝑑𝜙
0
3
𝜋/2 ∫ 4
2
𝜌 sin 𝜙 𝑑𝜌 𝑑𝜃 𝑑𝜙
0
3
𝜋/2
∫
sin 𝜙
0
𝜋/2
sin 𝜙
0
4
𝜌2 𝑑𝜌 𝑑𝜃 𝑑𝜙
3
4
𝜌3 𝑑𝜃 𝑑𝜙
3 3
∫
𝜋 37 𝜋/2
sin 𝜙 𝑑𝜃 𝑑𝜙
=
4 3 0
𝜋 37
= −
cos 𝜙∣𝜋0 2
4 3
37𝜋
=
.
12
□
3
Practice Problems
Practice problems from the recommended textbooks are:
∙ Stewart: Section 12.6, Exercises 1-9 odd, 17-27 odd; Section 12.7, Exercises 1-9 odd, 21-27
odd, 35
∙ Marsden/Tromba: Section 1.4, Exercise 1; Section 6.2, Exercises 13, 19, 23, 25
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